EXPERIMENT # 11 CALORIMETRY AND SPECIFIC HEAT PURPOSE ncasure the specific heat

Question

EXPERIMENT # 11 CALORIMETRY AND SPECIFIC HEAT PURPOSE ncasure the specific heat of a metal, and to use the results of this to estimate the mcts To mic weight, using the law of Dulong and Petit. INTRODUCTION lent is a form of energy, often referred to as thermal encrey, which flows from an object at a Heat temperature to one at a lower temperature when the two objects come in contact w temperat ith one er. The two objects will, after a short time, reach a state of thermal equilibrium, where another. The t are at the bicct wl equal the heat gained by the cooler one, a result of the law of conservation o where they same final temperature. If the system is well-insulated, the heat lost by the hotter same f object w This is the ga The heat of energY This is the basis for the calculation of the unknown metas specific heat. involved in a temperature change depends on three factors: the mass (m) of the object, e change (AT) the object undergoes and the specific heat (s) of the the substance changing one gram of the temperature ature. Specific heat is defined as the heat needed to raise the temperature one grarm by one Celsius degree. Its units, therefore, can be expressed as Jig-Cc. Water tempe ce by one C ific heat of 4.I84 J/g-C°. Heat, g. (in Joules) is calculated by the equation Water has a specific heat of Since the metal's specific heat is not known, we cannot calculate q for it, but we can y (gained by the water) g (lost by the metal) q = ms&T; calculate the ings, we can say that gained by the water. Assuming no heat loss to or gain by the surroundi where the subscripts refer to the metal, m and water, w. Finally, a historically important empirical law for estimating the atomic weight of a metal was discovered by Dulong and Petit many years ago. They found that, for many metals the product of the specific heat and the atomic weight is approximately 25. This experimental rule was one of the few tools early chemists had to estimate atomic weights. SA 25 where s is the specific heat in J/g-C and A is the atomic weight of the metal.

Explanation / Answer

q = m x s x T

Heat gained by cool water

85.4 x 4.184 x (24.4 - 21.3) = 1107.6 J

Heat lost by metal

Always Heat gained by cool water is equal to Heat lost by metal (Page 121)

hence heat lost by metal = 1107.6 J

Specific heat of metal

48.23 x Specific heat x (100-24.4) = 1107.6

Specific heat of metal = 0.303 J/g oC

Specific heat of metal 0.127 it may gold or lead

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